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In a topological space $X$, quoted from Wikipedia:

A point $x ∈ X$ is a cluster point of a sequence $(x_n)_{n ∈ N}$ if, for every neighbourhood $V$ of $x$, there are infinitely many natural numbers $n$ such that $x_n ∈ V$. If the space is sequential, this is equivalent to the assertion that $x$ is a limit of some subsequence of the sequence $(x_n)_{n ∈ N}$.

I was wondering when a topological space is not necessarily sequential, what is the relation between cluster point of a sequence and limit point of some subsequence of the sequence?

When the topological space is sequential space, why are the two equivalent?

Thanks and regards!

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Wikipedia is wrong: we need ($T_1$) Fréchet-Urysohn spaces for this to hold, also for the statement that there is a sequence from $S \setminus \{ p \}$ converging to $p$ if $p$ is a limit point of $S$. See my answer below. –  Henno Brandsma Feb 1 '12 at 10:06

1 Answer 1

In general, if $p$ is the limit of a subsequence $(x_{n_k})$ of the sequence $(x_n)$ in $X$ then $p$ is a cluster point of the sequence: let $O$ be a neighbourhood of $p$, then as $(x_{n_k})$ converges to $p$, there is some $K \in \mathbb{N}$ such that for all $k \ge K$ we have that $x_{n_k} \in O$, and $\{ n_k \mid k \ge K \}$ then are infinitely many natural numbers such that $x_{n_k}$ is in $O$.

For sequential spaces it does not to hold that a cluster point of a sequence is the limit of a subsequence of it. This will hold in $T_1$ Fréchet-Urysohn spaces (see below) and certainly in first countable spaces. The Arens space is a counterexample: it consists of all pairs $(n,m)$ in $\mathbb{N}^{+} \times \mathbb{N}^{+}$ (which are all isolated points), all $n \in \mathbb{N}^{+}$, which have basic neighbourhoods of the form $B(n,N) = \{ n \} \cup \{ (n,m) \mid m \ge N \}$, for $N \in \mathbb{N}^{+}$, so that $ ((n,m))_m \rightarrow n$ for all $n$, and the point $0$, which has basic neighbourhoods of the form that contain $0$, all but finitely many $n$ and for each of these $n$ a neighbourhood $B(n, N(n))$. One checks that this space is sequential and normal. Also, no sequence from $\mathbb{N}^{+} \times \mathbb{N}^{+}$ converges to $0$, but $0$ is in the closure of it. So if we index the set of all pairs as a sequence, $0$ is a cluster point of it, but not a subsequential limit.

For Fréchet-Urysohn spaces the set of all subsequential limits from a set $A$ equals the closure of $A$ (this is one of the definitions of being Fréchet-Urysohn), and if we have a sequence $(x_n)$ in a $T_1$ Fréchet-Urysohn space with a cluster point $p$, then either $p$ appears infinitely any times in the sequence, and this gives us a convergent subsequence to $p$, or we can assume $p$ does not appear at all in $A = \{ x_n \mid n \in \mathbb{N} \}$, and $p$ is in the closure of $A$, so is a subsequential limit from $A$, which allows us to define a subsequence from $(x_n)$ that converges to $p$ as well, using $T_1$ to get higher and higher indices from the sequence.

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+1, thanks! "This will hold in T1 Fréchet-Urysohn spaces (see below) and certainly in first countable spaces." I was wondering: (1) Is "T1 Fréchet-Urysohn spaces" a sufficient condition, while "first countable spaces" is a necessary condition? (2) Since Urysohn implies Fréchet, and T1 and Fréchet are the same thing, what does "T1 Fréchet-Urysohn spaces" mean? –  Tim Feb 2 '12 at 10:34
    
A first countable space is Fréchet-Urysohn, and is a sufficient condition. A $T_1$ Fréchet-Urysohn space also has the property (as I sketched in the last paragraph), but for first countable spaces I can prove the property (a cluster point is a subsequential limit) without the $T_1$ assumption. I'm not sure if I can omit it for the more general F-U spaces. A space is called Fréchet-Urysohn iff the sequential closure of any subset equals its closure. A space is sequential if sequentially closed sets are closed, a weaker property. NO relation to the separation axioms Fréchet and Urysohn –  Henno Brandsma Feb 2 '12 at 10:40
    
Thanks! From Wikipedia, "Two points x and y are separated if each of them has a neighbourhood that is not a neighbourhood of the other; that is, neither belongs to the other's closure." "X is T1, or accessible or Fréchet, if any two distinct points in X are separated." "X is T2½, or Urysohn, if any two distinct points in X are separated by closed neighbourhoods." So I think Urysohn implies Fréchet, and Fréchet and $T_1$ are the same thing. Isn't it? –  Tim Feb 2 '12 at 10:49
    
Sure, Fréchet is another name for $T_1$, and Urysohn implies it, and is sometimes called $T_{2\frac{1}{2}}$. But Fréchet-Urysohn is a sequence related property, also mentioned in the Wikipedia page for sequential spaces. It says that the closure equals the sequential closure. –  Henno Brandsma Feb 2 '12 at 21:12
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I'm not sure if there is a non-$T_1$ Fréchet-Urysohn space that has a sequence $(x_n)$ and a cluster point $p$ of that sequence without $p$ being a subsequential limit of it. That would show the necessity of $T_1$-ness in my (sketch of) proof that cluster points are subsequential limits in F-U spaces. –  Henno Brandsma Feb 3 '12 at 21:53

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